Problem: Factor the following expression: $4$ $x^2+$ $3$ $x$ $-1$
Solution: This expression is in the form ${A}x^2 + {B}x + {C}$ . You can factor it by grouping. First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(4)}{(-1)} &=& -4 \\ {a} + {b} &=& & & {3} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-4$ and add them together. Remember, since $-4$ is negative, one of the factors must be negative. The factors that add up to ${3}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${4}$ $ \begin{eqnarray} {ab} &=& ({-1})({4}) &=& -4 \\ {a} + {b} &=& {-1} + {4} &=& 3 \end{eqnarray} $ Next, rewrite the expression as ${A}x^2 + {a}x + {b}x + {C}$ $ {4}x^2 {-1}x +{4}x {-1} $ Group the terms so that there is a common factor in each group: $ ({4}x^2 {-1}x) + ({4}x {-1}) $ Factor out the common factors: $ x(4x - 1) + 1(4x - 1) $ Notice how $(4x - 1)$ has become a common factor. Factor this out to find the answer. $(4x - 1)(x + 1)$